I have an equation of a PI controller. What means here the velocity constant "R" exactly?



  • $\begingroup$ See, how beautiful formula I've made for you. :-) $\endgroup$
    – peterh
    Jun 28, 2016 at 8:53
  • $\begingroup$ Thank you very much. But can you explain me what it means? $\endgroup$ Jun 28, 2016 at 9:30
  • 1
    $\begingroup$ Yes, but not on the level of a professional. I've learned it in the University around 15 years ago. But it is a wonderful thing, control theory is much more wonderful as you could have ever imagined. Really, it is not about math. Really, it is not about machines. Really, it is about the life. The whole life is control theory. $\endgroup$
    – peterh
    Jun 28, 2016 at 10:07
  • $\begingroup$ If you don't get answer in some hours, ping me (write a comment beginning with "@peterh"), and I will explain what I can. $\endgroup$
    – peterh
    Jun 28, 2016 at 10:08
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    $\begingroup$ @gamlielbasha It seems that you are mixing the formula in the time domain ($t$) and in the Laplace/frequency domain ($s$). $\endgroup$
    – Karlo
    Jun 28, 2016 at 15:05

1 Answer 1


Quick Answer

The $R$ in the PI controller equation would best be described as a "weighting factor" that determines the influence of the integral term relative to the proportional term. The larger $R$ gets, the more the integrated error affects the output of the controller compared to the instantaneous error.

More explanation

Often PI controllers are expressed in the following form:

$u(t) = K e(t) + K_I \int e(t)dt$

where $K$ is called the proportional gain, and $K_I$ is called the integral gain. You tune the controller by changing $K$ and $K_I$ until the system behaves the way you want.

The equation given in the question expresses $K_I$ as a multiple of $K$, i.e. $K_I = RK$. Therefore if you substitute it back into the controller equation:

$u(t) = K e(t) + RK \int e(t)dt = K (e(t) + R \int e(t)dt)$

Now, instead of tuning $K$ and $K_I$ separately, you can look at the process as tuning the proportional gain $K$, then deciding how much influence the integral term has by adjusting $R$. This way of expressing the gains might be useful depending on how you approach the problem of tuning the controller. Really it's just mathematical semantics, you can use whichever form makes more sense to you.


$R$ has units of $\text{s}^{-1}$ (inverse seconds) in order to make the equation have consistent units. For example if $e(t)$ is a distance, then $\int e(t) dt$ has units of distance * time. Therefore, $R$ must have units of 1/time in order to cancel the time in the integral term.


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